Two soap bubbles of radii r_1 = 4 , cm and r_ | vedclass

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(B) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.Let $P

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$20$

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(B) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$,where $T$ is the surface tension of the soap solution.Let $P_1$ and $P_2$ be the excess pressures inside the bubbles of radii $r_1$ and $r_2$ respectively.$P_1 = \frac{4T}{r_1}$ and $P_2 = \frac{4T}{r_2}$.Since $r_1  P_2)$.The common surface will be curved towards the smaller bubble with a radius of curvature $r$.The pressure difference across the common surface is $\Delta P = P_1 - P_2 = \frac{4T}{r}$.Substituting the values: $\frac{4T}{r} = \frac{4T}{r_1} - \frac{4T}{r_2}$.$\frac{1}{r} = \frac{1}{r_1} - \frac{1}{r_2} = \frac{r_2 - r_1}{r_1 r_2}$.Therefore,$r = \frac{r_1 r_2}{r_2 - r_1} = \frac{4 	imes 5}{5 - 4} = 20 \, cm$.

Two soap bubbles of radii $r_1 = 4 \, cm$ and $r_2 = 5 \, cm$ are touching each other over a common surface $S_1S_2$ (as shown in the figure). The radius of curvature of this common surface will be ....... $cm$.

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